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Electrons of mass m with de-Broglie wavelength λ fall on the target in an X-ray tube. The cut-off wavelength (λ₀) of the emitted X-ray is
Options
(a) λ₀= 2mcλ²/h
(b) λ₀= 2h/mc
(c) λ₀= 2m²c²λ³/h²
(d) λ₀= λ
Correct Answer:
λ₀= 2mcλ²/h
Explanation:
λ = h / p ⇒ p = h / λ
KE of electrons = E = p² / 2m = h² / 2mλ²
Also in X-ray, λ₀ = hc / E = 2mcλ² / h
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Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The mirrors are inclined at an angle of 50⁰. The number of images formed for an object
- The phase difference between the flux linked with a coil rotating in a uniform
- A body of length 1 m having cross-sectional area 0.75 m² has heat flow through it
- If potential(in volts) in a region is expressed as V(x,y,z) = 6xy – y + 2yz,
- A stone falls freely under gravity. It covers distances h1, h2 and h3 in the first 5 secs
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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